Nature Reserve

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a forest that we model as a plane and live nn rare animals. Animal number ii has its lair in the point (xi,yi)(xi,yi). In order to protect them, a decision to build a nature reserve has been made.

The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.

For convenience, scientists have made a transformation of coordinates so that the river is defined by y=0y=0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.

Input

The first line contains one integer nn (1≤n≤1051≤n≤105) — the number of animals.

Each of the next nn lines contains two integers xixi, yiyi (−107≤xi,yi≤107−107≤xi,yi≤107) — the coordinates of the ii-th animal's lair. It is guaranteed that yi≠0yi≠0. No two lairs coincide.

Output

If the reserve cannot be built, print −1−1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10−610−6.

Formally, let your answer be aa, and the jury's answer be bb. Your answer is considered correct if |a−b|max(1,|b|)≤10−6|a−b|max(1,|b|)≤10−6.

Examples

input

Copy

1
0 1

output

Copy

0.5

input

Copy

3
0 1
0 2
0 -3

output

Copy

-1

input

Copy

2
0 1
1 1

output

Copy

0.625

Note

In the first sample it is optimal to build the reserve with the radius equal to 0.50.5 and the center in (0, 0.5)(0, 0.5).

In the second sample it is impossible to build a reserve.

In the third sample it is optimal to build the reserve with the radius equal to 5858 and the center in (12, 58)(12, 58).

題意：一個圓與X軸相切，問最小的半徑，使得圓包含所有給定的點。

解題思路：求最小，很容易想到二分。我們二分半徑，然後由於固定了與X軸相切，我們對於每一個點，就可以算出這個點在圓上的時候的圓心的極限距離。然後我們就可以求得每一個點的圓心的區間範圍。然後所有點的區間範圍都相交，那麼證明這個半徑可行，否則不可行。

#include<bits/stdc++.h>
using namespace std;
double x[100010],y[100010];
int N;
bool check(long double k)
{
    long double l=-100000000000000000.0,r=100000000000000000.0,t;
    for(int i=1;i<=N;i++)
    {
        if(y[i]>k*2)
            return 0;
        t=sqrt(k*k-(k-y[i])*(k-(y[i])));//計算圓心最遠的位置
        if(l < x[i]-t) l = x[i]-t;
        if(r > x[i]+t) r = x[i]+t;
    }
    return l<r;//所有圓心區間必須有交點
}
int main()
{

    scanf("%d",&N);
    for(int i=1;i<=N;i++)scanf("%lf %lf",&x[i],&y[i]);
    for(int i=1;i<=N;i++)
        if(y[i]*y[N]<0)
        {
            puts("-1");
            return 0;
        }
        else y[i]=y[i]>0?y[i]:-y[i];//必須都在一邊

    long double l=0,r=100000000000000000.0,m;
    for(int i=1;i<=100;i++)//二分半徑足夠多的次數
    {
        m=(l+r)/2.0;
        if(check(m))
            r=m;
        else
            l=m;
    }
    printf("%.10Lf",m);
    return 0;
}